A 26-foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at -2 feet per second (note that the rate is negative because the height is decreasing). At what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall?

Dx/dt  = 4,8 f/s

Explanation:

The ladder placed against a wall, and the ground formed a right triangle

with x and h the legs and L the hypothenuse

Then

L² = x² + h²          (1)

L = 26 f

Taking differentials on both sides of the equation we get

0  = 2x Dx/dt  + 2h Dh/dt    (1)

In this equation

x = 10   distance between the bottom of the ladder and the when we need to find, the rate of the ladder moving away from the wall

Dx/dt is the rate we are looking for

h = ?    The height of the ladder when  x = 10

As    L²  =  x²  + h²

h²  =  L²  -  x²

h²  =  (26)²  - (10)²

h²  =  676  -  100

h²  = 576

h = 24 f

Then equation (1)

0  = 2x Dx/dt  + 2h Dh/dt

2xDx/dt  = -  2h Dh/dt

10 Dx/dt  = - 24 ( -2 )      ( Note the movement of the ladder is downwards)

Dx/dt  =  48/10

Dx/dt  = 4,8 f/s

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